4418. November 8th: Dedekind cuts.18.1.Prove that the relation≤onDdefned in Defnition 18.2 is antisymmetric.Answer.Assume (A1|B1)≤(A2|B2) and (A2|B2)≤(A1|B1). By defnition, thismeansA1⊆A2andA2⊆A1. Since⊆is antisymmetric, this impliesA1=A2. Itthen Follows thatB2=Q±A2=A1=B1, proving that (A1|B1) = (A2|B2).±18.2.With notation as in the prooF oF Theorem 18.4, prove thatx<yFor allx∈Land ally∈M.Answer.Sincex∈L(and by defnition oF union over Families, Defnition 5.7)there exists an indexj∈Isuch thatx∈Aj. Sincey∈M,y∈BiFor alli∈I(bydefnition oF intersection over Families), and in particulary∈Bj. Since (Aj|Bj) is aDedekind cut, it Follows thatx < y, as required.±18.3.Provide a sensible defnition oF the ‘sum’ (A1|B1)+(A2|B2) oF two arbitraryDedekind cuts (A1|B1), (A2|B2). This should be done in such a way that iFq1,q2∈Qandι(q1) = (A1|B1),ι(q2)=(A2|B2), then (A1|B1A2|B2)=ι(q1+q2).
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